-k(-2-3)-(-2)(-k-5)=-2-(-2k+4)+-3

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Solution for -k(-2-3)-(-2)(-k-5)=-2-(-2k+4)+-3 equation: 



-k(-2-3)-(-2)(-k-5)=-2-(-2k+4)+-3

We move all terms to the left:

-k(-2-3)-(-2)(-k-5)-(-2-(-2k+4)+-3)=0

We add all the numbers together, and all the variables

-k(-5)-(-2)(-1k-5)-(-2-(-2k+4)+-3)=0

We use the square of the difference formula

-k(-5)-(-2)(-1k-5)-(-2-(-2k+4)-3)=0

We multiply parentheses

5k-(-2)(-1k-5)-(-2-(-2k+4)-3)=0

We multiply parentheses ..

5k-(+2k+10)-(-2-(-2k+4)-3)=0



We calculate terms in parentheses: -(-2-(-2k+4)-3), so:

-2-(-2k+4)-3

determiningTheFunctionDomain
-(-2k+4)-2-3

We add all the numbers together, and all the variables

-(-2k+4)-5

We get rid of parentheses

2k-4-5

We add all the numbers together, and all the variables

2k-9

Back to the equation:

-(2k-9)



We add all the numbers together, and all the variables

5k-(2k+10)-(2k-9)=0

We get rid of parentheses

5k-2k-2k-10+9=0

We add all the numbers together, and all the variables

k-1=0

We move all terms containing k to the left, all other terms to the right

k=1

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